Question

using mathematical induction prove that for any natural number 'n' the statement 4^2n > 15n is always true

Answer 1

we know, every natural numbers are the combination of even numbers and odd numbers .

case 1 :- Let's consider that n is an even number. e.g., n = 2r where n$\in\mathbb{I^+}$

$4^{2(2r)}>15(2r)$

or, $4^{4r}>30r$

or, $(4^4)^r>30$

hence, $(256)^r>30r$ this is correct statement. if we take r = 1
then it is clear that, 256 > 30 [ always true ]

case 2 :- let's consider that n is an odd number.
e.g., n = 2r + 1 , where n $\in\mathbb{I^+}$

$4^{2(2r+1)}>15(2r+1)$

or, $4^{4r+2}>30r+15$

or, $4^{4r}.4^2>30r+15$

or, $256^r.16>30r+15$

or, $16(256)^r>30r+15$ this statement is always true, for checking take r = 1
then, 16 × 256 > 30 + 15 [ always true ]

hence proved that for any natural number n the statement 4²ⁿ > 15n is always true.